2.6.3 Characteristic of Compound-wound motor
A compound-wound motor has both a series and a shunt field
winding, (i.e. one winding in series and one in parallel with the armature
circuit), by varying the number of turns on the series and shunt windings
and the directions of the magnetic fields produced by these windings
(assisting or opposing), families of (c/s) may be obtained to suit almost
all applications. There are two common types of compound motor
connection, the long-shunt connection and short-shunt connection. And
there are two different types of compound motors in common use, they
are the cumulative compound motor and the differential compound
motor. In the cumulative compound motor, the field produced by the
series winding aids the field produced by the shunt winding. The speed
of this motor falls more rapidly with increasing current than does that of
the shunt motor because the field increases. In the differential compound
motor, the flux from the series winding opposes the flux from the shunt
winding. The field flux, therefore, decreases with increasing load current.
Because the flux decreases, the speed may increases with increasing
load. Depending on the ratio of the series-to-shunt field ampere-turns,
the motor speed may increases very rapidly.
Fig.(2.15)
57
The torque-speed (c/s) of a cumulatively compound D.C motor
In the cumulative compounded D.C. motor, there is a component of
flux which is constant and anther component which is proportional to its
armature current (and thus to its load). Therefore, the cumulatively
compounded motor has a higher starting torque than a shunt motor
(whose flux is constant) but a lower starting torque than a series motor
(whose entire flux is proportional to armature current). At light loads, the
series field has a very small effect, so the motor behaves approximately
as a shunt D.C. motor. As the load gets very large, the series flux
becomes quite important and the torque-speed curve begins to look like a
series motor's (c/s). A comparison of the torque-speed (c/s) of each of
these type of machines is shown in figure (2.16).
The torque-speed (c/s) of a differentially compound D.C motor
In a differentially compound D.C. motor, the shunt magneto
motive force and series magneto motive force subtract from each other.
This means that as the load on the motor increases, Ia increases and the
flux in the motor decreases. But as the flux decreases, the speed of the
motor increases. This speed increases causes anther increases in load,
which further increases Ia , further decreasing the flux, and increasing the
speed again. The result is that a differentially compounded motor is
unstable and tends to run away. It is so bad that a differentially
compounded motor is unsuitable for any application.
Cumulatively
compound
n
n
Series
Shunt
T
T
Figure (2.16
58
2.7 D.C Motor Starter
If a D.C motor whose armature is stationary is switched directly to
its supply voltage, it is likely that the fuses protecting the motor will burn
out. Because the armature resistance is small, frequently being less than
one ohm. Thus, additional resistance must be added to the armature
circuit at the instant of closing the switch to start the motor.
As the speed of the motor increases. The armature conductors are
cutting flux and a generated voltage, acting in opposition to the applied
voltage, is produced, which limits the flow of armature current. Thus the
value of the additional armature resistance can then be reduced.
When at normal running speed, the generated e.m.f. is such that
no additional resistance is required in the armature circuit. To achieve
this varying resistance in the armature circuit on starting a D.C motor
starter is used, as shown in fig.(2.17). The starting handle is moved
slowly in a clockwise direction to start the motor. For a shunt-wound
motor, the field winding is connected to stud (1) or (M) via a sliding
contact on the starting handle. To give maximum field current hence
maximum flux, hence maximum torque on starting, since T ∝ φ.I a .
Fig.(2.17)
59
2.8 Speed Control of D.C Motor
2.8.1 Shunt-Wound Motor
The speed of a shunt-wound D.C motor, n, is proportional to
(V − I a .Ra )
φ
. The speed is varied either by varying the value of flux, ( φ ), or
by varying the value of ( Ra ). The former is achieved by using a variable
resistor in series with the field winding, as shown in fig.(2.18) and such a
resistor is called the shunt field regulator. As the value of resistance of
the shunt field regulator is increased, the value of the field current, ( I f ),
is decreased. This results in a decrease in the value of flux, ( φ ), and
1
hence an increase in the speed, since n ∝
φ
. Thus only speeds above that
given without a shunt field regulator can be obtained by this method.
Speeds below those given by
(V − I a .Ra )
φ
are obtained by increasing
the resistance in the armature circuit, as shown in fig.(2.18), where
n∝
V − I a (Ra + R )
φ
Since resistor (R) is in series with the armature, it carriers the full
armature current and results in a large power loss in large motors where a
considerable speed reduction is required for long periods.
Fig.(2.18)
60
2.8.2 Series-Wound Motor
The speed control of series-wound motors is achieved using
either (a) field resistance, or (b) armature resistance techniques.
(a) The speed of a D.C series-wound motor is given by :
⎛ V − IR ⎞
⎟⎟
n = K ⎜⎜
⎝ φ ⎠
Where (K) is a constant, (V) is the terminal voltage, ( R) is the
combined resistance of the armature and series field and ( φ ) is the
flux.
Thus, a reduction in flux results in an increase in speed. This is
achieved by putting a variable resistance in parallel with the field
winding and reducing the field current, and hence flux, for a given
value of supply current. A circuit diagram of this arrangement is
shown in fig. (2.19). A variable resistor connected in parallel with the
series-wound field to control speed is called a diverter speeds above
those given with no diverter are obtained by this method.
Fig.(2.19)
61
(b) speed below normal are obtained by connecting a variable
resistor is series with the field winding and armature circuit, as
shown in fig.(2.20). This effectively increases the value of (R) in
the equation.
⎛ V − I .R ⎞
⎟⎟
n = K ⎜⎜
⎝ φ ⎠
And thus reduces the speed. Since the additional resistor carries the
full supply current, a large power loss is associated with large motors
in which a considerable speed reduction is required for long periods.
Fig.(2.20)
Example (2.1) A D.C motor has a speed of (900 r.p.m) when connected
to a (460 V) supply. Find the approximate value of the speed of the motor
when connected to a (200 V) supply, assuming the flux decreases by
(30%) and neglecting the armature volt drop?
Solution:
E b1 = K φ1 .n1
E b 2 = K φ 2 .n 2
φ 2 = φ1 − φ1 × 0.3
φ 2 = 0.7φ1
Now
Eb1
φ × 900
= 1
Eb 2 0.7.φ1 × n2
n2 = 559 r.p.m
62
Example (2.2): A series motor has an armature resistance of (0.2 Ω) and
a series field resistance of (0.3 Ω). It is connected to a (240 V) supply and
at a particular load runs at (1440 r.p.m) when drawing (15 A) from the
supply.
(a) Determine the back e.m.f at this load.
(b) Calculate the speed of motor when the load is changed such that
the current is increased to (30 A). Assume that this cases a
doubling of flux.
Solution:
(a) at initial load, is given by
E b1 = V − I a (Ra + R f
)
Eb1 = 240 - 15(0.2+0.3)
= 232.5 Volt.
(b) When the current is increased to (30 A), the back e.m.f. is given
by.
E b 2 = V − I a (Ra + R f
)
=240 – 3o(0.2+0.3)
=225 volt
Now back e.m.f Eb ∝ φ .n
Thus
Eb1
φ .n
= 1 1
Eb 2 2.φ1 .n2
i.e.
232.5 φ1 × 1440
=
225
2 × φ1 × n2
n2 =
1440 × 225
= 696.77 r.p.m
232.5 × 2
********************************************
63
Example (2.3); A series motor runs at (800 r.p.m) when the voltage is
(400 V) and the current is (25 A). The armature resistance is (0.4 Ω) and
the series field resistance is (0.2 Ω). Determine the resistance to be
connected in series to reduce the speed to (600 r.p.m) with
same current.
Solution:
at (800 r.p.m)
E b1 = V − I ( Ra + R f )
= 400 – 25(0.4+0.2)
=385 volt
at (600 r.p.m), since the current is unchanged, the flux is unchanged.
Thus Eb ∝ φ × n ,
Eb 2 =
or
Eb ∝ n ,
and
Eb1 n1
=
Eb 2 n2
(385)(600) = 288.75 volt
(800)
And
E b 2 = V − I (Ra + R f + R )
288.75=400-25(0.4+0.2+R)
Rearranging gives
0 .6 + R =
400 − 288.75
= 4.45
25
From which, extra series resistance,
R=4.45-0.6
i.e. , R=3.85 Ω
thus the addition of a series resistance of (3.85 Ω) has reduced the speed
from (800 r.p.m) to (600 r.p.m).
*************************************************
64
Example (2.4): On full-load a (300 V) series motor takes (90 A) and runs
at (900 r.p.m) the armature resistance is (0.1 Ω) and the series winding
resistance is (50 mΩ). Determine the speed when developing full load
torque but with a (0.2 Ω) diverter in parallel with the field winding.
(assume that the flux is proportional to the field current).
Solution:
at (300 V)
E b1 = V − I (Ra + R f
)
IX
= 300 − 90(0.1 + 0.05)
=286.5 Volts
With the (0.2 Ω) diverter in parallel with ( R f )
The equivalent resistant
R=
0.2 × 0.05
= 0.04Ω
0.2 + 0.05
By current division, current I X = I ⎛⎜
0 .2
⎞
⎟
⎝ 0.2 + 0.05 ⎠
I X = 0.8I ,
I X = 0.8 I a 2
Torque, T ∝ I aφ and for full load torque
I a1φ1 = I a 2φ 2
Since flux is proportional to field current φ1 ∝ I a1 and φ 2 ∝ 0.8I a 2
Then (90)(90) = ( I a 2 )(0.8I a 2 )
,
I a22 =
(90)2 and
0 .8
I a 2 = 100.62 A
Hence Eb 2 = V − I a 2 ( Ra + R)
= 300 − 100.62(0.1 + 0.04) = 285.9 Volts
Back e.m.f. , Eb ∝ φ .n from which
new speed
n2 =
Eb1 φ1 × n1
I a1 × n1
=
=
Eb 2 φ 2 × n2 0.8 × I a 2 × n2
285.9 × 90 × 900
= 1004.4 r.p.m
286.5 × 0.8 × 100.62
65
2.9 The efficiency of a D.C. motor
It was stated in section (1.14), that the efficiency of a D.C. machine
is given by.
Efficiency, η =
output ⋅ power
× 100 0
0
input ⋅ power
Also, the total losses = I a2 Ra + I f V + C (for a shunt motor) and,
total losses= I 2 R + C (for a series motor), where C is the sum of the iron,
friction and windage losses, R is the total resistance for series motor
R = ( Ra + R f )
for a motor, the input power=VI
and the output power=VI-losses
hence,
⎛ VI − I a2 R − I f V − C ⎞
⎟ × 100 0
η =⎜
0
⎜
⎟
VI
⎝
⎠
(for shunt motor)
⎛ VI − IR − C ⎞
⎟ × 100 0 0
VI
⎝
⎠
η =⎜
(for series motor)
The efficiency of a motor is a maximum when the load is such that
I a2 Ra = I f V + C (for shunt motor), I 2 R = C (for series motor)
Example (2.5): A 250 V series motor draws a current of 40 A. The
armature resistance is 0.15 Ω and the field resistance is 0.05 Ω.
Determine the maximum efficiency of the motor.
Solution:
For series motor
⎛ VI − I 2 (Ra + R f ) − C ⎞
⎟ × 100
η =⎜
⎜
⎟
VI
⎝
⎠
⎛ VI − 2 I 2 ( Ra + R f ) ⎞
⎟ × 100
For maximum efficiency , η = ⎜⎜
⎟
VI
⎝
⎠
⎛ (250)(40) − 2(40) 2 (0.15 + 0.05) ⎞
⎟⎟ × 100 = 93.6 0
= ⎜⎜
0
(
250
)(
40
)
⎝
⎠
66
2.10 D.C Stepping Motors
D.C stepping motors are unique D.C motors that are used to
control automatic industrial processing equipment. D.C motors of this
type are fund in numerically controlled machines and robotic system
used by industry. They are very efficient and develop a high torque the
stepping motor is used primarily to change electrical pulses into a rotary
motion that can be used to produce mechanical movements.
The shaft of a D.C stepping motor rotates a specific number of
mechanical degrees with each incoming pulse of electrical energy. The
amount of rotary movement or angular displacement produced by each
pulse can be repeated precisely with each succeeding pulse from the drive
source. The resulting output of this device is used to accurately locate or
position automatic process machinery.
Fig.(2.20)
67
2.11 Electromechanical power control equipment
There are so many types of electromechanical power control
equipment used today that it is almost impossible to discuss each type.
However, some of the very important types will be discussed in the
following paragraphs.
2.11.1 Relays
Relays represent one of the most widely used control devices
available today. The electromagnet of a relay contains a stationary core.
Mounted close to one end of the core is a movable piece of magnetic
material called the armature. When the coil is activated electrically, it
produces a magnetic field in the metal core. The armature is then
attracted to the core, which in turn produces a mechanical motion. When
the coil is de-energized, the armature is returned to its original position by
spring action. Figure (2.21) shows a simplified diagram of the
construction of a relay that is used to control.
Common
Normal
Closed
(N.C)
Normal
Open
(N.O)
Fig.(2.21)
68
Relays use a small amount of current to create an electromagnetic
field that is strong enough to attract the armature. When the armature is
attracted, it either opens or closes the contacts. The contacts then either
turn (on) or (off) circuits that are using large amounts of current.
There are two types of contacts used in conjunction with most
relays. Normally open (N.O) and normally close (N.C). The (N.O)
contacts remain open when the relay coil is de-energized, and are closed
only when the relay is energized. The (N.C) contacts remain closed when
the relay is de-energized, and are open only when the coil is energized.
Application
D.C Motor Reversing
The direction of rotation of a permanent-magnet D.C motor can be
reversed by reversing the two power line as shown in fig.(2.22).
Forward
Reverse
Fig.(2.22)
69
2.11.2 Solenoids
A solenoid, shown in fig.(2.23) is an electromagnetic coil with a
movable core that is constructed of a magnetic. The core, or plunger, is
sometimes attached to an external spring. This spring causes the plunger
to remain in a fixed position until moved by the electromagnetic field that
is created by current through the coil. This external spring also causes the
core or plunger to return to its original position when the coil is
de-energized.
Solenoid are used for a variety of control applications. Many gas
and fuel oil furnaces use solenoid valves to automatically turn the fuel
supply (on) or (off) upon demand. Dishwashers used one or more
solenoids to control the flow of water.
Fig.(2.23)
70